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Quantum degeneracy problem, electron on a ring

  • #1

Homework Statement



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Homework Equations



Below

The Attempt at a Solution



So this is a lot like the infinite square well, except periodic. If S is an arc length, then [itex]S = \theta R[/itex] so [itex]\frac{d^2}{dS^2} = \frac{1}{R^2}\frac{d^2}{d\theta^2}[/itex], which is more convenient to use in the hamiltonian. So for the hamiltonian I get:

[tex]H = \frac{-\hbar^2}{2m}\frac{1}{R^2}\frac{d^2}{d\theta^2} + V_0[/tex]

With Schrodinger's equation, I get

[tex]\frac{d^2 \psi^2}{d\theta^2} = -\frac{2mR^2(E - V_0)}{\hbar^2}\psi[/tex]

Which gives solutions of the form [itex]\psi = \frac{1}{\sqrt{2\pi}}e^{\pm i k \theta}[/itex], where [itex]k = \sqrt{\frac{2mR^2(E - V_0)}{\hbar^2}}[/itex].

Then, because it's a ring, we need [itex]\psi(x + 2\pi) = \psi(x)[/itex] for any x, which gives us the requirement that k is an integer. So our energy levels are [itex]E = \hbar^2 k^2/2mR^2 + V_0[/itex], and it seems like they have a degeneracy of 4 because we have two functions for each k with the same energy, and then for each of them, the electron's spin can be up or down. Is that right?

As for part (b), I have no idea... Benzene has 6 free electrons, so according to my degeneracy, it completely fills the first energy level, and then there are 2 electrons in the 2nd energy level. Ok...then they ask about a compound with 4 electrons. This seems like it just fills the first energy level, but that seems too simple and stupid to be right.

Can anyone help me out?

Thanks!
 

Answers and Replies

  • #2
TSny
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k is an integer including zero. What is the degeneracy of the energy level corresponding to k = 0?
 
  • #3
k is an integer including zero. What is the degeneracy of the energy level corresponding to k = 0?
Well I guess that just has a degeneracy of 2 due to the spin, right?

Any idea on the aromatic thing? I'm totally clueless about that...
 
  • #4
Ah, I think I see, after what you just said and reading the wiki article on Huckel's Rule... The number of states for k =/= 0 is 4 for each energy level. For k = 0 it's 2. So like Huckel's Rule, for energy level n, there are 4n + 2 states. So benzene is aromatic because it has 6 electrons, so fully completes the n = 1 energy level. The other compound has 4 however, so it fills the n = 0 level but only half fills the n = 1 level, so it's not stable (though from the little I know of chemistry, I thought that just meant it's more reactive, not less stable).
 
  • #5
TSny
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I think your analysis is now correct. I'm not very clear on the reactive vs. stable interpretation either. Maybe someone can clarify it.
 

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